2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Multiply all terms by 12 to clear the denominators: \(4(2x - 5) - 3(x - 2) = 24\). Expand the brackets: \(8x - 20 - 3x + 6 = 24\). Simplify the left side: \(5x - 14 = 24\). Add 14 to both sides: \(5x = 38\). Divide by 5: \(x = 7.6\).

M1 for clearing fractions correctly, e.g., \(4(2x - 5) - 3(x - 2) = 24\) or better. A1 for \(7.6\) or equivalent fraction (e.g., \(\frac{38}{5}\) or \(7\frac{3}{5}\)).

Factorise the quadratic expression: \(2x^2 + 5x - 12 = (2x - 3)(x + 4) = 0\). Set each factor to zero: \(2x - 3 = 0 \implies x = 1.5\) and \(x + 4 = 0 \implies x = -4\).

M1 for correct factorisation \((2x - 3)(x + 4)\) or correct use of the quadratic formula. A1 for both solutions: \(1.5\) (or \(\frac{3}{2}\)) and \(-4\).

Use the area formula for a non-right-angled triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} \times 7 \times 9 \times \sin(42^\circ) = 31.5 \times 0.66913... = 21.0776...\text{ cm}^2\). Rounding to 3 significant figures gives \(21.1\text{ cm}^2\).

M1 for \(\frac{1}{2} \times 7 \times 9 \times \sin(42)\). A1 for \(21.1\) (accept answers in the range \([21.07, 21.1]\)).

Use the cosine rule: \(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(PQR)\). Substitute the given values: \(PR^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos(105^\circ) = 64 + 121 - 176 \times (-0.258819...) = 185 + 45.552... = 230.552...\). Take the square root: \(PR = \sqrt{230.552...} = 15.1839...\text{ cm}\). Rounding to 3 significant figures gives \(15.2\text{ cm}\).

M1 for a correct substitution into the Cosine Rule: \(8^2 + 11^2 - 2 \times 8 \times 11 \times \cos(105)\). A1 for \(15.2\) (accept answers in the range \([15.15, 15.25]\)).

The probability that the first ball is red is \(\frac{5}{8}\). Since the selection is without replacement, there are now 4 red balls left out of a total of 7 balls. The probability that the second ball is red given that the first was red is \(\frac{4}{7}\). The probability that both balls are red is \(\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\).

M1 for \(\frac{5}{8} \times \frac{4}{7}\). A1 for \(\frac{5}{14}\) or equivalent fraction, or \(0.357\) (accept \([0.357, 0.358]\)).

We are looking for the conditional probability \(P(\text{Chemistry} | \text{Physics})\). The number of students who study Physics is 18. Of these 18 students, 8 also study Chemistry. Therefore, the probability is \(\frac{8}{18} = \frac{4}{9}\).

M1 for a fraction with 18 in the denominator, or \(\frac{8/30}{18/30}\). A1 for \(\frac{4}{9}\) or equivalent fraction, or \(0.444\) (accept \([0.444, 0.445]\)).

The gradient of the line \(y = 3x - 5\) is 3. The gradient of the perpendicular line is the negative reciprocal: \(m = -\frac{1}{3}\). Using the point-slope form \(y - y_1 = m(x - x_1)\) with the point \((6, 2)\): \(y - 2 = -\frac{1}{3}(x - 6)\). Simplifying gives: \(y - 2 = -\frac{1}{3}x + 2 \implies y = -\frac{1}{3}x + 4\).

M1 for gradient of perpendicular line \(= -\frac{1}{3}\). A1 for \(y = -\frac{1}{3}x + 4\) or any equivalent fully simplified form (e.g., \(y = -0.333x + 4\)).

First, find the gradient of the given line by rearranging it into \(y = mx + c\) form: \(5y = -2x + 10 \implies y = -\frac{2}{5}x + 2\). The gradient of this line is \(-\frac{2}{5}\) (or \(-0.4\)). Since line \(L\) is perpendicular, its gradient is the negative reciprocal: \(m = -\frac{1}{-2/5} = \frac{5}{2} = 2.5\).

M1 for finding the gradient of the given line, e.g., \(-\frac{2}{5}\) or \(-0.4\). A1 for \(2.5\) or \(\frac{5}{2}\).

To solve the equation, first multiply all terms by $12$ (the lowest common multiple of the denominators) to clear the fractions:

$$4(x + 2) - 3(x - 1) = 24$$

Expand the brackets:

$$4x + 8 - 3x + 3 = 24$$

Simplify the left-hand side by combining like terms:

$$x + 11 = 24$$

Subtract $11$ from both sides of the equation:

$$x = 13$$

M1 for correctly clearing fractions to obtain a linear equation, e.g., $4(x+2) - 3(x-1) = 24$ (or equivalent)
A1 for $13$ (or $x = 13$)

By the sine rule:

$$\frac{\sin(\angle ABC)}{AC} = \frac{\sin(\angle ACB)}{AB}$$

Substitute the given values into the formula:

$$\frac{\sin(\angle ABC)}{11} = \frac{\sin(34^\circ)}{8}$$

$$\sin(\angle ABC) = \frac{11 \sin(34^\circ)}{8} \approx 0.76889$$

Calculate the inverse sine to find the acute angle:

$$\angle ABC = \sin^{-1}(0.76889) \approx 50.254^\circ$$

Rounding to $1$ decimal place gives $50.3^\circ$.

M1 for correct substitution into the sine rule, e.g., $\frac{\sin(\angle ABC)}{11} = \frac{\sin(34^\circ)}{8}$
A1 for $50.3$ (or $50.25$ to $50.3$)

Use the formula for conditional probability:

$$P(\text{Squash} | \text{Tennis}) = \frac{P(\text{Tennis and Squash})}{P(\text{Tennis})}$$

Substitute the given probabilities:

$$P(\text{Squash} | \text{Tennis}) = \frac{0.12}{0.4} = 0.3$$

M1 for writing $\frac{0.12}{0.4}$ or for a correct conditional probability expression
A1 for $0.3$ (or equivalent fraction, e.g., $\frac{3}{10}$)

The gradient of the line $y = 3x - 5$ is $3$. Since the lines are perpendicular, the gradient, $m$, of the perpendicular line is the negative reciprocal:

$$m = -\frac{1}{3}$$

Use the equation of a straight line, $y - y_1 = m(x - x_1)$, with the point $(6, 2)$:

$$y - 2 = -\frac{1}{3}(x - 6)$$

$$y - 2 = -\frac{1}{3}x + 2$$

$$y = -\frac{1}{3}x + 4$$

M1 for identifying the perpendicular gradient as $-\frac{1}{3}$ (or $-0.333$)
A1 for $y = -\frac{1}{3}x + 4$ or equivalent (e.g., $y = -0.333x + 4$)

To eliminate $y$, multiply the second equation by $2$:

$$10x - 2y = 22$$

Now, add this new equation to the first equation:

$$(3x + 2y) + (10x - 2y) = 17 + 22$$

$$13x = 39$$

$$x = 3$$

Substitute $x = 3$ back into the second equation to find $y$:

$$5(3) - y = 11$$

$$15 - y = 11$$

$$y = 4$$

Thus, $x = 3$ and $y = 4$.

M1 for a correct attempt to eliminate one variable (e.g., multiplying the second equation by 2 and adding, or writing $y = 5x - 11$ and substituting)
A1 for $x = 3, y = 4$ (both correct)

Use the formula for the area of a non-right-angled triangle:

$$\text{Area} = \frac{1}{2} a b \sin C$$

Substitute the given values into the formula:

$$45 = \frac{1}{2} \times 9 \times 14 \times \sin(\angle PQR)$$

$$45 = 63 \sin(\angle PQR)$$

$$\sin(\angle PQR) = \frac{45}{63} = \frac{5}{7}$$

Calculate the inverse sine to find the acute angle:

$$\angle PQR = \sin^{-1}\left(\frac{5}{7}\right) \approx 45.58^\circ$$

Rounding to $1$ decimal place gives $45.6^\circ$.

M1 for setting up the area equation, e.g., $45 = \frac{1}{2} \times 9 \times 14 \times \sin(\angle PQR)$ or $\sin(\angle PQR) = \frac{5}{7}$
A1 for $45.6$ (or $45.58\dots$)

First, find the midpoint, $M$, of the line segment $AB$:

$$M = \left(\frac{2+8}{2}, \frac{5+1}{2}\right) = (5, 3)$$

Next, calculate the gradient of the line segment $AB$:

$$m_{AB} = \frac{1-5}{8-2} = \frac{-4}{6} = -\frac{2}{3}$$

The gradient, $m_{\perp}$, of the perpendicular bisector is the negative reciprocal of $m_{AB}$:

$$m_{\perp} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} = 1.5$$

Now, find the equation of the perpendicular line passing through the midpoint $(5, 3)$:

$$y - 3 = 1.5(x - 5)$$

$$y - 3 = 1.5x - 7.5$$

$$y = 1.5x - 4.5$$

M1 for finding either the midpoint $(5, 3)$ or the perpendicular gradient $1.5$ (or $\frac{3}{2}$)
A1 for $y = 1.5x - 4.5$ or $y = \frac{3}{2}x - \frac{9}{2}$

First, multiply the entire equation by the common denominator \(x(x + 3)\):

\(10(x + 3) - 6x = x(x + 3)\)

Expand both sides:

\(10x + 30 - 6x = x^2 + 3x\)

\(4x + 30 = x^2 + 3x\)

Rearrange the equation into standard quadratic form \(ax^2 + bx + c = 0\):

\(x^2 - x - 30 = 0\)

Factorise the quadratic expression:

\((x - 6)(x + 5) = 0\)

This gives the solutions:

\(x = 6\) or \(x = -5\)

M1: For multiplying through by the common denominator to obtain \(10(x + 3) - 6x = x(x + 3)\) or equivalent.
M1: For expanding and simplifying to a three-term quadratic equation, e.g., \(x^2 - x - 30 = 0\).
M1: For factorising their quadratic into two brackets, e.g., \((x - 6)(x + 5) = 0\), or correct substitution into the quadratic formula.
A1: For both correct solutions \(x = 6\) and \(x = -5\).

The total length of the garden and the path combined is \(12 + 2w\) metres.
The total width of the garden and the path combined is \(8 + 2w\) metres.

The total combined area is given by:

\((12 + 2w)(8 + 2w) = 165\)

Expand the brackets:

\(96 + 24w + 16w + 4w^2 = 165\)

\(4w^2 + 40w + 96 = 165\)

Rearrange into a standard quadratic form:

\(4w^2 + 40w - 69 = 0\)

Solve using the quadratic formula where \(a = 4\), \(b = 40\), and \(c = -69\):

\(w = \frac{-40 \pm \sqrt{40^2 - 4(4)(-69)}}{2(4)}\)

\(w = \frac{-40 \pm \sqrt{2704}}{8}\)

\(w = \frac{-40 \pm 52}{8}\)

This gives two potential solutions:

\(w = \frac{12}{8} = 1.5\) or \(w = \frac{-92}{8} = -11.5\)

Since the width \(w\) must be positive, we discard the negative solution.

Thus, \(w = 1.5\).

M1: For writing a correct expression for the combined dimensions, i.e., \((12 + 2w)\) and \((8 + 2w)\).
M1: For setting up the area equation: \((12 + 2w)(8 + 2w) = 165\).
A1: For expanding and simplifying to the quadratic equation \(4w^2 + 40w - 69 = 0\).
A1: For solving the quadratic equation to get \(w = 1.5\) (accepting the rejection of the negative root).

Substitute the expression for \(y\) from the first equation into the second equation:

\(x^2 + x(2x - 3) = 6\)

Expand the term \(x(2x - 3)\):

\(x^2 + 2x^2 - 3x = 6\)

Combine like terms to form a quadratic equation:

\(3x^2 - 3x - 6 = 0\)

Divide the entire equation by 3 to simplify:

\(x^2 - x - 2 = 0\)

Factorise the quadratic:

\((x - 2)(x + 1) = 0\)

This gives the x-values:

\(x = 2\) or \(x = -1\)

Now, substitute these x-values back into the linear equation \(y = 2x - 3\) to find the corresponding y-values:

For \(x = 2\):

\(y = 2(2) - 3 = 1\)

For \(x = -1\):

\(y = 2(-1) - 3 = -5\)

So the solutions are \(x = 2, y = 1\) and \(x = -1, y = -5\).

M1: For substituting \(y = 2x - 3\) into the second equation to get \(x^2 + x(2x - 3) = 6\).
A1: For simplifying to \(3x^2 - 3x - 6 = 0\) (or \(x^2 - x - 2 = 0\)).
M1: For solving their quadratic equation to find two values of \(x\) (\(x = 2\) and \(x = -1\)).
A1: For finding both corresponding \(y\) values correctly: \(y = 1\) when \(x = 2\) and \(y = -5\) when \(x = -1\) (or writing as coordinate pairs \((2, 1)\) and \((-1, -5)\)).

First, use the Cosine Rule to find the length of side \(AC\):

\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\)

\(AC^2 = 7^2 + 11^2 - 2 \cdot 7 \cdot 11 \cdot \cos(115^\circ)\)

\(AC^2 = 49 + 121 - 154 \cdot (-0.422618)\)

\(AC^2 = 170 + 65.0832 = 235.0832\)

\(AC = \sqrt{235.0832} \approx 15.332\text{ cm}\)

Next, apply the Sine Rule to find the angle \(BAC\) (let's call it \(A\)):

\(\frac{\sin A}{BC} = \frac{\sin(115^\circ)}{AC}\)

\(\frac{\sin A}{11} = \frac{\sin(115^\circ)}{15.332}\)

\(\sin A = 11 \cdot \frac{0.906308}{15.332} \approx 0.65022\)

\(A = \arcsin(0.65022) \approx 40.56^\circ\)

To 1 decimal place, angle \(BAC = 40.6^\circ\).

M1: For applying the Cosine Rule: \(AC^2 = 7^2 + 11^2 - 2(7)(11)\cos(115^\circ)\).
A1: For calculating \(AC \approx 15.3\text{ cm}\) (or more precise, e.g., \(15.332\)).
M1: For applying the Sine Rule: \(\frac{\sin(BAC)}{11} = \frac{\sin(115^\circ)}{15.33}\) (or using the Cosine Rule to find angle \(BAC\)).
A1: For angle \(BAC = 40.6^\circ\) (or \(40.6\)).

Let's first determine the interior angle \(PQR\) of triangle \(PQR\).

The bearing from \(P\) to \(Q\) is \(060^\circ\). Draw a North-South line at \(Q\). The angle from the South direction at \(Q\) pointing back to \(P\) is also \(060^\circ\) (alternate angles).

The bearing of \(R\) from \(Q\) is \(130^\circ\). The angle between the North line and the line \(QR\) is \(130^\circ\). The angle between the South direction and \(QR\) is \(180^\circ - 130^\circ = 50^\circ\).

Therefore, the interior angle \(PQR = 60^\circ + 50^\circ = 110^\circ\).

Now, we apply the Cosine Rule to find the distance \(PR\):

\(PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(PQR)\)

\(PR^2 = 15^2 + 22^2 - 2 \cdot 15 \cdot 22 \cdot \cos(110^\circ)\)

\(PR^2 = 225 + 484 - 660 \cdot (-0.34202)\)

\(PR^2 = 709 + 225.733 = 934.733\)

\(PR = \sqrt{934.733} \approx 30.57\text{ km}\)

To 3 significant figures, the distance is \(30.6\text{ km}\).

M1: For calculating the interior angle \(PQR = 110^\circ\).
M1: For applying the Cosine Rule: \(PR^2 = 15^2 + 22^2 - 2(15)(22)\cos(110^\circ)\).
A1: For evaluating \(PR^2 \approx 934.7\) (or \(935\)).
A1: For \(PR = 30.6\text{ km}\) (accept \(30.6\)).

First, find the angle \(XYZ\) (let's call it \(Y\)) using the area formula:

\(\text{Area} = \frac{1}{2} \cdot XY \cdot YZ \cdot \sin Y\)

\(40.5 = \frac{1}{2} \cdot 8.5 \cdot 12.4 \cdot \sin Y\)

\(40.5 = 52.7 \cdot \sin Y\)

\(\sin Y = \frac{40.5}{52.7} \approx 0.7685\)

Since angle \(Y\) is obtuse (greater than \(90^\circ\)):

\(Y = 180^\circ - \arcsin(0.7685) \approx 180^\circ - 50.22^\circ = 129.78^\circ\)

Now, use the Cosine Rule to find the length of \(XZ\):

\(XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos Y\)

\(XZ^2 = 8.5^2 + 12.4^2 - 2 \cdot 8.5 \cdot 12.4 \cdot \cos(129.78^\circ)\)

\(XZ^2 = 72.25 + 153.76 - 210.8 \cdot (-0.63984)\)

\(XZ^2 = 226.01 + 134.88 = 360.89\)

\(XZ = \sqrt{360.89} \approx 19.00\text{ cm}\)

To 3 significant figures, the length of \(XZ\) is \(19.0\text{ cm}\).

M1: For writing a correct equation for the area: \(0.5 \times 8.5 \times 12.4 \times \sin(Y) = 40.5\).
A1: For calculating the obtuse angle \(Y \approx 129.8^\circ\) (or \(129.78^\circ\)).
M1: For applying the Cosine Rule: \(XZ^2 = 8.5^2 + 12.4^2 - 2(8.5)(12.4)\cos(129.8^\circ)\).
A1: For \(XZ = 19.0\text{ cm}\) (or \(19\)).

Let \(R\) be the event that it rains, and \(F\) be the event that the weather station forecasts rain.

We are given:

\(P(R) = 0.3\)

\(P(R') = 0.7\)

\(P(F|R) = 0.9\)

\(P(F|R') = 0.2\)

We want to find the conditional probability \(P(R|F)\):

\(P(R|F) = \frac{P(R \cap F)}{P(F)}\)

First, find the probability of both rain and a forecast of rain:

\(P(R \cap F) = P(R) \cdot P(F|R) = 0.3 \cdot 0.9 = 0.27\)

Next, find the probability of no rain and a forecast of rain:

\(P(R' \cap F) = P(R') \cdot P(F|R') = 0.7 \cdot 0.2 = 0.14\)

Then, the total probability that the station forecasts rain is:

\(P(F) = P(R \cap F) + P(R' \cap F) = 0.27 + 0.14 = 0.41\)

Now, calculate \(P(R|F)\):

\(P(R|F) = \frac{0.27}{0.41} = \frac{27}{41} \approx 0.6585\)

To 3 significant figures, the probability is \(0.659\).

M1: For calculating \(P(\text{Rain} \cap \text{Forecast}) = 0.3 \times 0.9 = 0.27\).
M1: For calculating \(P(\text{No Rain} \cap \text{Forecast}) = 0.7 \times 0.2 = 0.14\).
A1: For the total probability of forecasting rain \(P(F) = 0.41\).
A1: For finding the conditional probability \(\frac{27}{41}\) or \(0.659\) (accept answers rounding to \(0.659\)).

First, find the midpoint \(M\) of the line segment \(AB\):

\(M = \left(\frac{-3 + 5}{2}, \frac{8 + 2}{2}\right) = (1, 5)\)

Second, find the gradient \(m_{AB}\) of the line segment \(AB\):

\(m_{AB} = \frac{2 - 8}{5 - (-3)} = \frac{-6}{8} = -\frac{3}{4}\)

Third, the gradient of the perpendicular bisector is the negative reciprocal of \(m_{AB}\):

\(m_{\perp} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}\)

Fourth, write the equation of the line passing through \(M(1, 5)\) with gradient \(\frac{4}{3}\):

\(y - 5 = \frac{4}{3}(x - 1)\)

Multiply the entire equation by 3 to clear the fraction:

\(3(y - 5) = 4(x - 1)\)

\(3y - 15 = 4x - 4\)

Rearrange the equation into the form \(ay + bx = c\):

\(3y - 4x = 11\)

M1: For finding the midpoint of \(AB\) as \((1, 5)\).
M1: For finding the gradient of \(AB\) as \(-\frac{3}{4}\).
M1: For finding the gradient of the perpendicular line as \(\frac{4}{3}\).
A1: For the final equation \(3y - 4x = 11\) or any equivalent integer equation in the form \(ay + bx = c\) (e.g., \(4x - 3y = -11\)).

First, find the angle \(ABC\) using bearings: \(\angle ABC = 215^\circ - 115^\circ = 100^\circ\). Next, apply the cosine rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substitute the values: \(AC^2 = 110^2 + 145^2 - 2 \cdot 110 \cdot 145 \cdot \cos(100^\circ)\). Simplify the expression: \(AC^2 = 12100 + 21025 - 31900 \cdot (-0.173648) = 33125 + 5539.38 = 38664.38\). Solve for \(AC\): \(AC = \sqrt{38664.38} \approx 196.63\text{ m}\). Giving the answer to 3 significant figures yields \(197\text{ m}\).

M1 for finding angle \(ABC = 100^\circ\) M1 for correct substitution into the cosine rule, e.g. \(110^2 + 145^2 - 2 \cdot 110 \cdot 145 \cdot \cos(100^\circ)\) A1 for intermediate evaluation \(38664\dots\) or \(\sqrt{38664\dots}\) A1 for final answer \(197\) or \(196.6\dots\)

First, find the midpoint of the line segment \(AB\): \(\text{Midpoint } M = \left(\frac{2+6}{2}, \frac{-3+5}{2}\right) = (4, 1)\). Next, find the gradient of the line \(AB\): \(m_{AB} = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\). The gradient of the perpendicular line is the negative reciprocal: \(m_{\perp} = -\frac{1}{2} = -0.5\). Now, use the gradient-intercept form \(y = mx + c\) with the midpoint \((4, 1)\) and the perpendicular gradient \(-0.5\): \(1 = -0.5(4) + c \Rightarrow 1 = -2 + c \Rightarrow c = 3\). Thus, the equation of the perpendicular bisector is \(y = -0.5x + 3\).

M1 for finding the gradient of \(AB\) as \(2\) M1 for finding the perpendicular gradient as \(-0.5\) (or negative reciprocal of their gradient) M1 for finding the midpoint of \(AB\) as \((4, 1)\) A1 for the correct equation \(y = -0.5x + 3\) or equivalent form \(y = -\frac{1}{2}x + 3\)

(a) Using the Cosine Rule in \(\triangle ABD\):
\(BD^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(\angle BAD)\)
\(BD^2 = 85^2 + 60^2 - 2 \cdot 85 \cdot 60 \cdot \cos(110^\circ)\)
\(BD^2 = 7225 + 3600 - 10200 \cdot (-0.34202)\)
\(BD^2 = 10825 + 3488.6 = 14313.6\)
\(BD = \sqrt{14313.6} = 119.64\text{ m}\).
Rounded to the nearest metre, \(BD = 120\text{ m}\).

(b) In \(\triangle BCD\), the sum of angles is \(180^\circ\).
\(\angle BDC = 180^\circ - (42^\circ + 58^\circ) = 80^\circ\).
Using the Sine Rule:
\(\frac{CD}{\sin(58^\circ)} = \frac{BD}{\sin(42^\circ)}
Using the more accurate value \)BD = 119.64\text{ m}\):
\(CD = \frac{119.64 \cdot \sin(58^\circ)}{\sin(42^\circ)} = \frac{119.64 \cdot 0.84805}{0.66913} = 151.62\text{ m}\).
To 3 significant figures, \(CD = 152\text{ m}\).
(If using the given \(BD = 120\text{ m}\):
\(CD = \frac{120 \cdot \sin(58^\circ)}{\sin(42^\circ)} = 152.08 \approx 152\text{ m}\)).

(c) Total Area = \(\text{Area of } \triangle ABD + \text{Area of } \triangle BCD\)
\(\text{Area of } \triangle ABD = \frac{1}{2} \cdot AB \cdot AD \cdot \sin(110^\circ) = \frac{1}{2} \cdot 85 \cdot 60 \cdot \sin(110^\circ) = 2396.2\text{ m}^2\).
\(\text{Area of } \triangle BCD = \frac{1}{2} \cdot BD \cdot CD \cdot \sin(\angle BDC) = \frac{1}{2} \cdot 119.64 \cdot 151.62 \cdot \sin(80^\circ) = 8932.6\text{ m}^2\).
Total Area = \(2396.2 + 8932.6 = 11328.8 \approx 11300\text{ m}^2\) (to 3 s.f.).
(If using \(BD = 120\text{ m}\) and \(CD = 152\text{ m}\):
\(\text{Area of } \triangle BCD = \frac{1}{2} \cdot 120 \cdot 152 \cdot \sin(80^\circ) = 8981.5\text{ m}^2\).
Total Area = \(2396.2 + 8981.5 = 11377.7 \approx 11400\text{ m}^2\) (to 3 s.f.)). All values in range \([11300, 11400]\) are correct and accepted.

(d) The shortest distance from \(A\) to the line \(BD\) is the perpendicular height, \(h\), of \(\triangle ABD\) with base \(BD\).
\(\text{Area of } \triangle ABD = \frac{1}{2} \cdot BD \cdot h\)
Using the accurate value \(BD = 119.64\text{ m}\):
\(2396.2 = \frac{1}{2} \cdot 119.64 \cdot h \implies h = \frac{2 \cdot 2396.2}{119.64} = 40.05 \approx 40.1\text{ m}\).
(If using the given value \(BD = 120\text{ m}\):
\(2396.2 = \frac{1}{2} \cdot 120 \cdot h \implies h = \frac{2 \cdot 2396.2}{120} = 39.94 \approx 39.9\text{ m}\).
Alternatively, using right-angled trigonometry:
\(\sin(\angle ADB) = \frac{85 \cdot \sin(110^\circ)}{119.64} \implies \angle ADB = 41.87^\circ\).
\(h = 60 \cdot \sin(41.87^\circ) = 40.05 \approx 40.1\text{ m}\)). All values in range \([39.9, 40.1]\) are accepted.

(a) [4 marks]
M1: For correct substitution into the Cosine Rule: \(85^2 + 60^2 - 2 \cdot 85 \cdot 60 \cdot \cos(110)\).
A1: For correct evaluation of parts: \(7225 + 3600 - (-3488.6)\) or \(14313.6\).
A1: For \(BD = 119.64...\)
A1: For final rounding statement concluding that \(BD = 120\text{ m}\) to the nearest metre.

(b) [4 marks]
M1: For finding \(\angle BDC = 180 - 42 - 58 = 80^\circ\).
M1: For correct substitution into the Sine Rule: \(\frac{CD}{\sin(58)} = \frac{BD}{\sin(42)}\).
A1: For rearranging to find \(CD\) using either \(119.64\) or \(120\).
A1: For \(CD = 152\text{ m}\) (accept range \([151.6, 152.1]\)).

(c) [4 marks]
M1: For correct method to find the area of \(\triangle ABD\), e.g., \(\frac{1}{2} \cdot 85 \cdot 60 \cdot \sin(110)\) (or \(2396\)).
M1: For correct method to find the area of \(\triangle BCD\), e.g., \(\frac{1}{2} \cdot BD \cdot CD \cdot \sin(80)\).
A1: For one correct triangle area calculated: \(2396\) or \(8933\) (or \(8982\)).
A1: For total area in the range \([11300, 11400]\).

(d) [3 marks]
M1: For recognizing that the shortest distance is the perpendicular height of \(\triangle ABD\).
M1: For setting up an equation, e.g., \(\frac{1}{2} \cdot 119.64 \cdot h = 2396\) or \(h = 60 \cdot \sin(41.87)\).
A1: For answer in the range \([39.9, 40.1]\).

(a)(i) Time taken from A to B = \(\frac{180}{x}\) hours.

(a)(ii) Time taken from B to A = \(\frac{180}{x-15}\) hours.

(b) Since the total time is \(7.5\) hours:
\(\frac{180}{x} + \frac{180}{x-15} = 7.5\)
Divide both sides by \(7.5\):
\(\frac{24}{x} + \frac{24}{x-15} = 1\)
Multiply by the common denominator, \(x(x - 15)\):
\(24(x - 15) + 24x = x(x - 15)\)
\(24x - 360 + 24x = x^2 - 15x\)
\(48x - 360 = x^2 - 15x\)
Rearrange to form a quadratic equation:
\(x^2 - 15x - 48x + 360 = 0\)
\(x^2 - 63x + 360 = 0\) (as required).

(c) Using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{-(-63) \pm \sqrt{(-63)^2 - 4 \cdot 1 \cdot 360}}{2 \cdot 1}\)
\(x = \frac{63 \pm \sqrt{3969 - 1440}}{2}\)
\(x = \frac{63 \pm \sqrt{2529}}{2}\)
Using \(\sqrt{2529} \approx 50.28916\):
\(x_1 = \frac{63 + 50.28916}{2} = 56.64458 \approx 56.64\)
\(x_2 = \frac{63 - 50.28916}{2} = 6.35542 \approx 6.36\)
So, \(x = 56.64\) or \(x = 6.36\).

(d) For the return journey speed to be positive, we must have \(x - 15 > 0\), which implies \(x > 15\).
Thus, we reject \(x = 6.36\) and select \(x = 56.64\) km/h.
Return speed = \(56.64 - 15 = 41.64\text{ km/h}\).
Time taken for return journey = \(\frac{180}{41.64} = 4.32276\text{ hours}\).
Converting the fractional part to minutes:
\(0.32276 \times 60 = 19.36\text{ minutes}\).
Therefore, the time taken is 4 hours and 19 minutes (to the nearest minute).

(a)(i) [1 mark]
B1: For \(\frac{180}{x}\).

(a)(ii) [1 mark]
B1: For \(\frac{180}{x-15}\).

(b) [5 marks]
M1: For setting up the sum of times equation: \(\frac{180}{x} + \frac{180}{x-15} = 7.5\).
M1: For removing algebraic fractions by multiplying by the common denominator \(x(x-15)\).
M1: For expanding brackets correctly, e.g., \(24(x-15)\) as \(24x - 360\) or \(180(x-15)\) as \(180x - 2700\).
A1: For a correct intermediate step, e.g., \(48x - 360 = x^2 - 15x\) or \(360x - 2700 = 7.5x^2 - 112.5x\).
A1: For fully correct algebraic manipulation leading to \(x^2 - 63x + 360 = 0\) with no errors seen.

(c) [4 marks]
M1: For identifying coefficients and substituting into the quadratic formula (allow 1 sign error).
A1: For finding the discriminant term correctly: \(\sqrt{2529}\) or \(50.29\).
A1: For \(x = 56.64\).
A1: For \(x = 6.36\).

(d) [4 marks]
M1: For rejecting \(x = 6.36\) with a valid reason, e.g., \(x - 15 > 0\), or selecting \(x = 56.64\).
M1: For calculating the return journey speed or time, e.g., \(56.64 - 15 = 41.64\) or \(\frac{180}{41.64}\).
A1: For \(4.32...\) hours.
A1: For converting to \(4\text{ hours } 19\text{ minutes}\) (accept \(4\text{ h } 19\text{ m}\)).